find the solutions to the equation x^2+2x+5=0?

Question

a. x= -1+-2i
b. x= 1+-2i
c. x= -2+-4i
d. x= 2+-4i

Answer 1

Quadratic formula

x = - b ± √b² - 4ac / 2a

x² + 2x + 5 = 0

let

a = 1

b = 2

c = 5

x = - 2 ± √(2)² - 4(1)(5) / 2(1)

x = - 2 ± √4 - 20 / 2

x = - 2 ± √- 16 / 2. .no real numbers solution

- - - - - - - - - - -

Imaginary number solution

x = - 2 ± i √16 / 2

x = - 2 ± 4i / 2

- - - - - -

Solvint for +

x = - 2 + 4i / 2

x = - 1 + 2i

- - - - - - -

Solving for -

x = - 2 - 4i / 2

x = - 1 - 2i

- - - - - - - -s-

Answer 2

First, rearrange the equation to equivalent 0: 0 = x^2 +2x +(ok-5) = a^2 + bx + c. the quadratic formula then states that x = -b+- sqrt (b^2 - 4ac)/ (2a), the place c is okay-5. by way of fact 3 is a answer to the equation, then x = 3, a = a million, b = -2, c = ok-5, then the only variable to unravel for in the quadratic equation would be, ok. algebraic answer: (-((x)(2*a)+b)^2 +b^2)/(a*a million) + 5 = ok. as quickly as values are substituted: (-((3)(2*a million)+2)^2 + 2^2)/(4*a million) +5 = ok. I purely rearranged the formula to unravel for ok. the respond for ok is -10. which might simplify the above equation to x^2 +2x -15 = 0. ***the answer to that's ok = -10 and x = 3, -5. (x = 3 became the given answer).

Answer 3

It is clearly a, based upon the quadratic formula:

ax^2 + bx + c = 0
a = 1, b = 2, c = 5

[-b ± sqr(b^2 - 4ac)]/2a = x

Notice the negative inside of the square root gives you the i.

Answer 4

x^2 + 2x + 5 = 0

x^2 + 2x + 1 + 4 = 0

x^2 + 2x + 1 = -4

(x + 1)^2 = -4

x + 1 = +/- 2i
x = -1 +/- 2i

Answer 5

[-2 +/- sqrt(2^2-4*1*5)]/2=
[-2+/- sqrt(4-20)]/2=
[-2+/- 4i]/2=
-1+/- 2i for the answer is a.

Answer 6

Use the quadratic formula:

(-2 +/- sqrt( 4 - 20 ) ) / 2 =
(-2 +/- sqrt( -16 ) ) / 2 =
(-2 +/- sqrt( 16 ) * sqrt( -1 ) ) / 2 =
-1 +/- 2 * sqrt(-1)

i = sqrt(-1)

-1 +/- 2i

a is the answer :)

Answer 7

x^2+2x+5=0
(x^2+2x+1)+4=0
(x+1)^2+4=0
(x+1)^2=-4
x+1=-2i
x=-2i-1
so a is correct

Answer 8

x = [- 2 ± √(-16] / 2
x = [- 2 ± 4i ] / 2
x = -1 ± 2i
ANSWER a

Answer 9

It's a!

(-b (+or-) sqrt(b^2-4ac))/2a

(-2(+or-)(sqrt -16))/2

(-2(+or-)4i)/2

-1(+or-)2i